Solve {l}{x+y+z=22}{3x+y=47}{x-4z=2}

$$\left\{ \begin{array} { l } { x + y + z = 22 } \\ { 3 x + y = 47 } \\ { x - 4 z = 2 } \end{array} \right.$$

Solve for x, y, z

$$x=14$$

$$y=5$$

$$z=3$$

Short Steps Using Substitution

Solve x+y+z=22 for x.

$$x=-y-z+22$$

Substitute -y-z+22 for x in the second and third equation.

$$3\left(-y-z+22\right)+y=47 -y-z+22-4z=2$$

Solve these equations for y and z respectively.

$$y=-\frac{3}{2}z+\frac{19}{2} z=-\frac{1}{5}y+4$$

Substitute -\frac{3}{2}z+\frac{19}{2} for y in the equation z=-\frac{1}{5}y+4.

$$z=-\frac{1}{5}\left(-\frac{3}{2}z+\frac{19}{2}\right)+4$$

Solve z=-\frac{1}{5}\left(-\frac{3}{2}z+\frac{19}{2}\right)+4 for z.

$$z=3$$

Substitute 3 for z in the equation y=-\frac{3}{2}z+\frac{19}{2}.

$$y=-\frac{3}{2}\times 3+\frac{19}{2}$$

Calculate y from y=-\frac{3}{2}\times 3+\frac{19}{2}.

$$y=5$$

Substitute 5 for y and 3 for z in the equation x=-y-z+22.

$$x=-5-3+22$$

Calculate x from x=-5-3+22.

$$x=14$$

The system is now solved.

$$x=14 y=5 z=3$$